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Olin [163]
3 years ago
14

3. If the net work done on an object is negative, then the object's kinetic energy

Physics
1 answer:
g100num [7]3 years ago
4 0

Answer:

I think the answer is B I hope this helps also Im sorry If I'm wrong

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Your toaster has a power cord with a resistance of 2.2x10^−2 Ω connected in series with a 9.7 Ω nichrome heating element. The po
kati45 [8]

Answer:

(A) 654.545 Kw

(B) 2.885\times 10^5kw

Explanation:

We have given resistance of the toaster R_1=2.2\times 10^{-2}ohm=0.022ohm

Resistance of nichrome heating element R_2=9.7ohm

Both the resistances are connected in series so same current will flow through the circuit

Potential difference across the toaster V = 120 volt

So current i=\frac{120}{0.022}=5454.5454A

(a) Power dissipated in toaster P=i^2R=5454.5454^2\times 0.022=654545.441W=654.545kw

(B) Power dissipated in heating element  P=i^2R=5454.5454^2\times 9.7=2.885\times 10^8W=2.885\times 10^5kw

5 0
3 years ago
To practice Problem-Solving Strategy 21.1 Coulomb's Law. Three charged particles are placed at each of three corners of an equil
Salsk061 [2.6K]

Answer:

The force felt by charge 3 is F=(-5.6*10⁻⁶,3.36⁻⁵)N

Explanation:

As the superposition principle applies to static charges, we can find the net electric force as the sum of the two forces felt by q3.

Looking at the drawing and knowing that they form an equilateral triangle of lenght 4 we can conclude that each internal angle is 60°.

So, the positions in our coordinate system are:

r_1=(0,0)\\r_2=(4\ cos(60\°),4\ sin(60\°))\\r_3=(4,0)\\

Now  using Coulomb's force:

F_{ij}=\frac{-kq_iq_j}{d^2}(\vv{r}_j-\vv{r}_i)

Where d=4, q1 = -7.8*10⁻⁹C, q2 = -15.6 *10⁻⁹C, q3 = 8.0 *10⁻⁹C, k=8.98*10⁹, e0=8.8*10¹⁰:

Replacing we get 2 equations:

F_{13}=\frac{-kq_1q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(-0.04\ cos(60\°),-0.04\ sin(60\°))\\\\F_{23}=\frac{-kq_2q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(0.04-0.04\ cos(60\°),-0.04\ sin(60\°))\\

To work with the sam

F=∑F_i=3.5*10⁻⁴(0.023,0.032)+7*10⁻⁴(-0.016,0.032)=

=((3.5*10⁻⁴-7*10⁻⁴)*0.016,(3.5*10⁻⁴+7*10⁻⁴)*0.032)=

F=(-5.6*10⁻⁶,3.36⁻⁵)N

7 0
3 years ago
A system of two objects has ΔKtot = 6 J and ΔUint = -5 J. Part A How much work is done by interaction forces? Express your answe
Elina [12.6K]

A) +5 J

B) +1 J

Explanation:

A)

The internal forces (interaction forces) acting on a system do not change the mechanical energy (sum of potential and kinetic energy) of the system.

However, these forces are responsible for converting the energy from one form into another; the work done by these forces is equal to the amount of energy converted from one form into the other.

In this problem, we have:

\Delta U=-5 J is the loss in potential energy of the system

\Delta K=+6 J is the gain in kinetic energy of the system

By looking at these numbers, this means that the internal forces have converted 5 J of energy from potential energy into kinetic energy (while the additional +1 J missing is due to external forces, as explained in part B).

Therefore, the work done by internal forces is

W = +5 J

B)

First of all, we calculate the change in mechanical energy of the system.

The mechanical energy of a system is the sum of its kinetic energy (K) and its potential energy (U):

E=K+U

So, the change in mechanical energy is equal to the sum of the changes of kinetic energy and the changes of potential energy:

\Delta E= \Delta K + \Delta U

In this problem:

\Delta K=+6 J

\Delta U=-5 J

So, the change in mechanical energy is:

\Delta E=+6+(-5)=+1 J

According to the work-energy theorem, the work done by external forces on a system is equal to the change in mechanical energy of the system: therefore in this case, the work done by external forces is

W=\Delta E=+1 J

5 0
3 years ago
As best you can, describe how we can tell the age of the Earth and rocks found on it.
Viktor [21]

Answer:

We can determine the age of the Earth and the rocks found on Earth with techniques such as measuring ice cores and digging and analysing fossils.

Explanation:

Hopefully this helped!

4 0
2 years ago
The half-life of a certain isotope is 12 years. How much of a 600 g sample will remain after 36 years?
miskamm [114]

36/12= 3   3 IS THE AMOUT OF HALFLIVES

ONE HALF LIFE = 300

TWO HALF LIFE = 150

THREE HALF LIFE = 75


*Divide 600 three times

8 0
3 years ago
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